<span><span>Составьте</span><span>уравнение касательной</span><span>к графику функции</span><span>f</span><span>(</span><span>x</span><span>)</span><span>=</span><span>x</span><span>^ 2</span><span>в</span><span>точке с</span><span>абсциссой</span><span>х0</span><span>= 1.</span></span>
<span><span>y = f(x0) + f'(x0)(x-x0)</span></span>
<span><span>f(1) = 1^2 = 1</span></span>
<span><span>f'(x)=2x</span></span>
<span><span>f'(1) = 2*1=2</span></span>
<span><span>В итоге имеем: y = 1+2(x-1) = 2x -2+1 = 2x - 1</span></span>
81×10в 3-й степени или 8,1×10в 4-ой степени
а)x(x^4-1)+2(2^3-1)
б)b(a+c)-a(a+c)=(a+c)(b-a)